Post #2400400

In fact P3 rhomb tilings have been proven to be _three_-colourable. This is also true of any edge-to-edge plane tiling with parallelograms: a bit of reasoning about angle sums shows that the border of any finite patch must include a tile with at most 2 neighbours, so you can inductively 3-colour the rest of the patch and then you're guaranteed to be able to pick a colour for that last tile. And then, again, Erdős–de Bruijn says a three-colouring of the whole plane must exist. But again that proof doesn't come with any algorithm for delivering a specific three-colouring known to extend to the whole plane. I tried to use the same brute-force approach to generate a three-coloured P3 substitution system, by running multiple passes of my refiner starting from the orientation-classified version of P3. No luck: after two refinement passes the graph still had chromatic number 4, and after three, it became too big to wait for Sage to try to colour it at all. I'd _like_ a 3-coloured refinement of the P3 substitution system, but I haven't yet found a way to construct one, by this technique or any other!